Home  18.013A  Chapter 0  Section 0.4 


Binomial coefficients count the number of subsets of an n element set having k elements in them. The Stirling number here counts the number of partitions of a set of n elements into k disjoint blocks. Prove these two statements.
Solution:
The subsets of an n element set of size k are of two kinds.
Those which do not contain the nth element: these are actually k element subsets of an the n1 element set obtained by ignoring n.
and those that contain n: these all have k1 elements of the remaining n1.
This gives the recursion
C(n, k) = C(n1, k) + C(n1, k1)
which is what is computed in the first spreadsheets.
A partition of an n element set into k blocks can also be of either of two kinds. In one the element n is all by itself in a block; and we have a partition of the rest into k1 blocks.
Otherwise, it comes from a partition of the rest into k blocks, and then it can go into any one of them; so there are k * (the number of partitions of an n1 element set into k blocks) different ways that this second possibility can happen.
This gives the recursion used in the spreadsheet here, which is
S(n, k) = S(n1 ,k1) + k*S(n1, k)
which can be read as "n stands alone, or with one of the k blocks that exist without it".
